How would one find the largest integer $n$ such that $n! < a^n$ for some integer $a$?

Question

How would one find the largest $n$ that satisfies

$$n!<a^n$$ where $a,n\in\mathbb{Z}^+$ for a given $a$ analytically?

Answer 1

A good approximation can be obtained from Stirling's formula. Taking the $n$th root of $$ a^n\approx n^ne^{-n}\sqrt{2\pi n}$$ we get $a\approx n/e$ or $$ n\approx ea.$$ For example, with $a=1000$ we get the estimate $n\approx 2718$ when the correct value is $n=2713$.

Answer 2

Trivially we have $$n!<n^n$$ while Stirling’s approximation gives $$ n!< (1+o(1))\sqrt{2\pi n}\left(\frac{n}{e}\right)^n,$$ but for $n>2\pi$ you could say something like $$ n!<e\left(\frac{n}{e}\right)^{n+1}.$$